If the wavelength of the first line of the Paschen series of hydrogen atom is 720 nm, then the wavelength of the second line of this series is _________ nm. (Nearest integer)
Detailed Step-by-Step Solution
This problem uses the Rydberg Equation for the hydrogen emission spectrum. Instead of calculating the exact value of the Rydberg constant (\(R_H\)), taking a ratio of the two equations makes the math much simpler.
Step 1: Formula and 1st Line of Paschen Series
The Rydberg formula is given by:
For the Paschen series, electrons fall to the 3rd shell, so \( n_1 = 3 \).
The first line occurs when the electron transitions from \( n_2 = 4 \). We are given \( \lambda_1 = 720 \text{ nm} \).
\( \frac{1}{720} = R_H \left( \frac{1}{9} - \frac{1}{16} \right) = R_H \left( \frac{16 - 9}{144} \right) \)
\( \frac{1}{720} = R_H \left( \frac{7}{144} \right) \quad \text{--- (Equation 1)} \)
Step 2: Set up equation for the 2nd Line
The second line of the Paschen series occurs when the electron transitions from \( n_2 = 5 \) to \( n_1 = 3 \). Let its wavelength be \( \lambda_2 \).
\( \frac{1}{\lambda_2} = R_H \left( \frac{1}{9} - \frac{1}{25} \right) = R_H \left( \frac{25 - 9}{225} \right) \)
\( \frac{1}{\lambda_2} = R_H \left( \frac{16}{225} \right) \quad \text{--- (Equation 2)} \)
Step 3: Take the ratio to solve for \(\lambda_2\)
Dividing Equation 1 by Equation 2 cancels out \( R_H \):
\( \frac{\lambda_2}{720} = \frac{7}{144} \times \frac{225}{16} \)
Now, isolate \( \lambda_2 \):
Notice that \( \frac{720}{144} = 5 \), so we can simplify the expression easily:
\( \lambda_2 = 35 \times \frac{225}{16} \)
\( \lambda_2 = \frac{7875}{16} = 492.1875 \text{ nm} \)
Conclusion: The calculated wavelength is 492.1875 nm. Rounding off to the nearest integer gives us 492.
Master Hydrogen Emission Spectra
The hydrogen emission spectrum (Lyman, Balmer, Paschen, Brackett, and Pfund series) is one of the most frequently tested topics in JEE Main and NEET. Solving by taking the ratio of states is a time-saving trick that prevents rounding errors caused by manually calculating the Rydberg constant. For a complete conceptual breakdown, read our comprehensive notes on the Structure of Atom Class 11 Chemistry.
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